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3 years ago

Will it give me the right answers

Chemistry

1 answer:

iren2701 [21]3 years ago

6 0

Answer:

erijjqeujkf

Explanation:

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3 years ago

Need help !!!!! ASAP

Korolek [52]

<h2>Hello!</h2>

The answer is:

The new volume is 2.84 L.

$V_{2}=2.84L$

To solve the problem, we need to remember what STP means. STP means that the gas is at standard temperature and pressure, or 273.15 K (0°C) and 1 atm.

Also, we need to use the Combined Gas Law, since the temperature, the pressure and the volume are being changed.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using , Gay-Lussac's Law, Charles's Law, and Boyle's Law.

The law is defined by the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

So, we are given the following information:

$V_{1}=2L\\P_{1}=1atm\\T_{1}=0\°C=273.15K\\P_{2}=80kPa=0.8atm\\T_{2}=37\°C=37+273=310K$

Then, isolating the new volume, and substituting, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\V_{2}=\frac{P_{1}V{1}}{T_{1}}*\frac{T_{2}}{P_{2}}\\\\V_{2}=\frac{1atm*2L}{273.15K}*\frac{310K}{0.8kPa}=2.84L$

Hence, the new volume is 2.84 L.

$V_{2}=2.84L$

Have a nice day!

4 0

4 years ago